- #1

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_{n+1}, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1?

Thank you in advance

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- Thread starter shanepitts
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- #1

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Thank you in advance

- #2

Mark44

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In aBelow is a screen shot of a solution to a problem. The part I don't fathom is after the ratio test is applied to the denominator. How can, noting that a_{n+1}, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1?

Thank you in advance

View attachment 83898

BTW, when you post a question here,

- #3

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In a_{n}, the denominator is the product ##1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)##. What will be the next factor in the denominator for a_{n + 1}?

BTW, when you post a question here,please don't delete the three parts of the homework template.

Thanks for the quick reply

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?

- #4

Mark44

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It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?Thanks for the quick reply

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?

- #5

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It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?

You get 2n+1 instead of the (2n-1)(2n+1), no?

- #6

Mark44

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You get (2n + 1). The (2n - 1) factor is the one from aYou get 2n+1 instead of the (2n-1)(2n+1), no?

- #7

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You get (2n + 1). The (2n - 1) factor is the one from a_{n}.

Thanks,

I'm sorry but I still don't fathom.

Let's say

α

Why the extra 2n-1 in the α

- #8

Mark44

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This is not aThanks,

I'm sorry but I still don't fathom.

Let's say

α_{n}=2n-1 then α_{n+1}should equal 2(n+1)-1=2n+1

$$a_n = \frac{x^n}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}$$

Now, what is a

You need to ask yourself how many factors are in the denominator of a

shanepitts said:Why the extra 2n-1 in the α_{n+1}?

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